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$3SATplus$

  • Input: 2 CNF formulas $F_1$, $F_2$ where all clauses have exactly $3$ literals.

  • Question: Does every truth assignment satisfy at-least as many $F_2$ clauses as $F_1$'s?

Assume $3SATPLUS \in NP$

Prove $3SAT \leq_p$ $3SATplus$

The reduction is as follows

$F_1 = (x \lor \bar{x} \lor x) \land \dots \land (x \lor \bar{x} \lor x)$

For $n$ clauses. This means it will always accept $n$ clauses.

$F_2 = F$

This is polytime because its $2n$ clauses only $n$ more than $F$

if $F \in 3SAT$, then $F_1 = F_2$ since they both satisfy $n$ clauses so $F_1, F_2 \in 3SATplus$

I'll prove the contrapositive

If $F \notin 3SAT$, then $F_1$ satisfies $n$ clauses and $F_2$ can satisfies only up to $n-1$ so $F_1 > F_2$ always. Hence, $F_1, F_2 \notin 3SATplus$

Is there anything else left to say or does this confirm its $NP-hard$?

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bob
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  • It seems odd to assume that $3SATPLUS\in NP$ because that probably isn't true. – Tom van der Zanden Jul 14 '19 at 08:38
  • The problem, as stated, is that every truth assignment must satisfy at least as many $F_2$ clauses as $F_1$ clauses. A certificate that demonstrates this is true for a single assignment is not valid. The complement problem is "there is at least one truth assignment which satisfies more $F_1$ clauses" is in $NP$, and thus this problem is in $coNP$. – Tom van der Zanden Jul 14 '19 at 11:40

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If $F\in 3SAT$ it means that there is some assignment that satisfies $n$ clauses. It doesn't mean that every assignment satisfies all clauses. This makes the reduction wrong, since there might be an assignment for $F\in 3SAT$ that satisfies fewer than $n$ clauses. The reduction you've proposed would be valid if you started with $Tautology$, but then, that problem isn't known to be $NP$-complete.

Your problem is $coNP$-complete (as the reduction from $Tautology$ shows), so finding a reduction from $3SAT$ would prove $NP\subseteq coNP$.

Tom van der Zanden
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